MSD Module 1 Design of Valve Gear Dr. F. Tauheed Valve gear mechanism is a subassembly of IC engine and its function is to open and close the inlet and exhaust valve at a proper time with respect to the position of piston and crankshaft. Fuel is admitted into the cylinder when engine valve is open and the burnt gas escapes when exhaust valve is open. Main parts of valve gear mechanism: 1. Valve 2. Spring 3. Tappet 4. Rocker arm 5. Push rod 6. Cam 7. Camshaft

Camshaft is rotated by means of a belt drive from the crankshaft. As the camshaft rotates, the cam pushes the follower and push rod upward. The rocker arm is pivoted at its centre by means of a fulcrum pin. When the right end of the rocker arm is pushed up by push rod, the left end moves downward. This compresses the spring and valve rod down in the cylinder, causing valve to open. When the follower moves over the circular portion of the spring expands and closes the valve. The spring pushes the left end of the rocker arm upward. This causes the right end to move downward and keeps the follower in contact with the cam. One end of the spring is secured to the valve stem and the other end rests on the cylinder. The spring is initially compressed. The force due to initial compression presses the valve down on its seat. Opening of the valve is due to oscillatory motion of the rocker arm. Design of Valve (i) Diameter of valve port: Let, a  area of piston v  mean velocity of piston

a p  area of port

v p  mean velocity of gas flowing through the port

Then, a pv p  av  and a  d 2 d  diameter of the port p 4 p p

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N Mean velocity of the piston v  2l p 60 where, l  length of stroke (m) N  engine speed (rpm)

The allowable mean velocities of the gas vp based on type of engine can be found in data book. (ii) Diameter of the valve head: For a seat angle 450, the projected width of the valve seat is given by the following empirical relation w  0.05 to 0.07d p

Diameter of the valve head dv  d p  2w

(iii) Thickness of valve disc: Valve disk is a circular disk freely supported around its periphery and subjected to uniformly distributed pressure pm ax . Thickness of the disk is calculated as

pmax t  kdp  b Where k  constant ( k  0.42 for steel and k  0.54 for cast iron) 2  b  permissible bending stress (N/mm ) = 50 – 60 N/mm2 for carbon steel = 100 – 120 N/mm2 for alloy steel Thickness of valve disk at the edges = 0.75 to 0.85t. (iv) Diameter of the valve stem:

d p  d p  Diameter of valve stem ds    6.35 to  11  8   8  The valve is subjected to spring force when seated. The spring force is assumed to be concentrated force at the centre. In this position the stress in the valve is given by 1.4P  2d  s s P  t  2 1  s = spring force in N t  3d p 

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(v) Maximum valve lift: hma x = maximum valve lift For continuous flow, the mean velocity of the gas flow through the valve is equal to that through the opening of the valve when it is lifted to a maximum. Therefore, maximum lift of the valve is obtained by equating area across the valve to the area of the port.  d d h cos   d 2  h  p p max 4 p max 4cos d For flat headed valve h  p as   0 ma x 4 The valve has the same angle as the valve seating surface. In practice, the angle of valve seat is made 0.50 to 10 more than the valve angle. This is called interference angle and helps in effective sealing. ------Problem: Design exhaust valve for a horizontal diesel engine using following data: Cylinder bore = 150 mm Length of stroke = 275 mm Engine speed = 500 rpm maximum gas pressure of 3.5MPa seat angle = 450 calculate 1. diameter of the valve port 2. Diameter of valve head 3. thickness of the valve head 4. stem diameter and 3. maximum lift of the valve. Solution: 0 Given D 150mm , l  275mm , N  500rpm, pmax  3.5MPa ,   45 Step I: Diaameter of the valve port  Area of the piston a  1502 mm2 4 N 500 Mean velocity of the piston v  2l  2275 m/s p 60 60

From data book assume v p  50 m/s a pv p  av

 d p  45.41mm say 46mm ------(i) Step II: Diameter of valve head For a seat angle 450, the projected width of the valve seat is given by the following empirical relation w  0.05 to 0.07d p

Taking w  0.06d p  2.76mm say 3mm

Diameter of the valve head dv  d p  2w  52mm ------(ii) (iii) Thickness of valve disc:

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pmax t  kdp  5.11mm say 5.5 mm ------(iii)  b Taking k  0.42 2  b  50 N/mm (iv) Diameter of the valve stem:

d p  d p  Diameter of valve stem ds    6.35 to  11  8   8  = 12.1mm to 16.75 mm

ds 15mm ------(iv) (v) Maximum valve lift: hmax = maximum valve lift d h  p 16.26mm max 4cos ------

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